Is ($x^2$ - 1) contained in ker($\phi$)?

Question

Let $\phi$₁: $\mathbb{Z}$[x] -> $\mathbb{Z}$ be the evaluation homomorphism at 1. I know that ker($\phi$) = < x - 1 > but would ($x^2$ - 1) $\in$ ker($\phi$)? Thank you in advance.

Answer 1

Yes. Both because $x^2-1$ is a multiple of $x-1$ and thus in the ideal $\langle x-1\rangle$, and because the evaluation-at-$1$ map sends $x^2-1$ to $1^1-1=0$ and thus $x^2-1$ is in the kernel of the map.